Question

how do I graph y^2+4x=0 using the
(x-h)^2=4p(y-k)

Answer 1

What do you mean exactly? Are you asking how you can put the first equation into the form you've seen for the second?

Answer 2

yes i need to graph it and find the focus the directrix and p

Answer 3

i would really appreciate the help I'm new haha

Answer 4

Okay, just a sec...doing multiple things at once ;)

Answer 5

I can be patient

Answer 6

Okay...

Answer 7

Your first and second equations are both parabolas, but they're different. The second equation will look like ones you're used to, whereas the first will open up differently.
\[y^2+4x=0 \rightarrow y^2-4x\](i.e. subtracting 4x from both sides) puts your parabola in a form where you can read off things like focus and directrix. I'll continue in a sec.

Answer 8

So, I made a mistake...great!

It should be,
\[y^2+4x=0 \rightarrow y^2=-4x\]

Answer 9

Now, do me a favor and go to this website:

www.wolframalpha.com

and in the box at the beginning, type in:

y^2+4x=0

Look at the graph and then come back. I'll continue with the focus and directrix.

Answer 10

:) ty

Answer 11

i think its suppose to be x= -y^2/4(1)

Answer 12

because the focus is (1,0)
and the directirx is (-1,0)

Answer 13

but thank u for your help soo much it was lovely ;)

Answer 14

The parabola you have is one that opens on its side. The form of this kind of parabola is\[(y-k)^2=4a(x-h)\]where a is your focus and (h,k) is the vertex.

In your case, both h and k are zero, and if you compare the form with your equation, you'll see,\[y^2=4ax \rightarrow a=-1\] so your focus is at

(-1,0)

and directrix

(1,0)

Answer 15

hmm you have given me a lot to think about you have the opposite opinion of someone else i asked hmm ok ty :)

Answer 16

You're welcome.

Your equation above, x=-y^2/4(1) is the same as what I gave you...just looks different.

Answer 17

thank you so much i really appreciate it :D